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In a six-digit integer N F(k) is the value of the k-th

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In a six-digit integer N F(k) is the value of the k-th [#permalink]

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New post 29 Aug 2008, 00:28
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In a six-digit integer N F(k) is the value of the k-th digit. For example, F(4) is the value of the hundreds digit of N. Is N divisible by 7?

1. F(1) = F(4), F(2) = F(5), F(3) = F(6)
2. F(1) = F(2) = ... = F(6)

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New post 29 Aug 2008, 00:49
should be D.

Not sure though
I am confident that B is sufficient on its own
2) 111111 is divisible by 7 and any other number will be x* 111111 ,where x is a digit other than 0.

1) The numbers i tried worked.
345345, 123123, 567567. will try to find whats the logic used here.

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New post 29 Aug 2008, 00:50
yes a+gamma ... the OA is D. good try... but will post the resoning later on. till then u can try

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Re: something different.. [#permalink]

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New post 29 Aug 2008, 01:05
N = x x x x x x

F(k) = value of the kth digit.

From ( ii )

A B A B A B
will be divisible by 7.

From ( ii )

A A A A A A

will be divisible by 7


Thus answer should be "D"

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Re: something different.. [#permalink]

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New post 29 Aug 2008, 07:21
arjtryarjtry wrote:
In a six-digit integer N F(k) is the value of the k-th digit. For example, F(4) is the value of the hundreds digit of N. Is N divisible by 7?

1. F(1) = F(4), F(2) = F(5), F(3) = F(6)
2. F(1) = F(2) = ... = F(6)


ABABAB
1) 100000*A +10000*B+ 1000*C +100*A + 10B +C
= 1001C +10010B+1001000A = 1001 (100A+10B+C)
= (100A+10B+C) * 1001
1001 --> is divisiable by 7

To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number

100-2 --> 98 is divisable by 7
Suffcieint
2) A* 111111

111111 -->11111-2 =11109
11109 -->1110-18 = 1092 =
1092 --> 105 is divisable by 7

sufficient

Any other short cuts.
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Last edited by x2suresh on 29 Aug 2008, 08:41, edited 1 time in total.

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Re: something different.. [#permalink]

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New post 29 Aug 2008, 07:24
x2suresh wrote:
arjtryarjtry wrote:
In a six-digit integer N F(k) is the value of the k-th digit. For example, F(4) is the value of the hundreds digit of N. Is N divisible by 7?

1. F(1) = F(4), F(2) = F(5), F(3) = F(6)
2. F(1) = F(2) = ... = F(6)


ABABAB
1) 100000*A +10000*B+ 1000*A +100*B + 10A +B
= 101010*A+10101* B
= (2a+b) * 10101
10101 --> is divisiable by 7

To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number

1010-2 --> 108 is divisable by 7
Suffcieint
2) A* 111111

111111 -->11111-2 =11109
11109 -->1110-18 = 1092 =
1092 --> 105 is divisable by 7

sufficient

Any other short cuts.


Hey guys,

I found another rule. (googled). this looks faster way for this problem

To know if a number is a multiple of seven or not, we can use also
3 coefficients (1 , 2 , 3). We multiply the first number starting
from the ones place by 1, then the second from the right by 3,
the third by 2, the fourth by -1, the fifth by -3, the sixth by -2,
and the seventh by 1, and so forth.


ABCABC --> 0 which is divisable by 7
AAAAAA -- > 0 which is divisable by 7
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Last edited by x2suresh on 29 Aug 2008, 08:35, edited 1 time in total.

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Re: something different.. [#permalink]

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New post 29 Aug 2008, 08:30
x2suresh wrote:
arjtryarjtry wrote:
In a six-digit integer N F(k) is the value of the k-th digit. For example, F(4) is the value of the hundreds digit of N. Is N divisible by 7?

1. F(1) = F(4), F(2) = F(5), F(3) = F(6)
2. F(1) = F(2) = ... = F(6)


ABABAB
1) 100000*A +10000*B+ 1000*A +100*B + 10A +B
= 101010*A+10101* B
= (2a+b) * 10101
10101 --> is divisiable by 7

To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number

1010-2 --> 108 is divisable by 7
Suffcieint
2) A* 111111

111111 -->11111-2 =11109
11109 -->1110-18 = 1092 =
1092 --> 105 is divisable by 7

sufficient

Any other short cuts.


Am I reading this wrong? Is the number ABABAB or ABCABC? Obviously, your method would still be able to test:

ABCABC = 100000A + 10000B + 1000C + 100A + 10B + C = 100100A + 10010B + 1001C
= 1001*(100A + 10B + C)

1001/7 = 143 --> sufficient

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Re: something different.. [#permalink]

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New post 29 Aug 2008, 08:42
zoinnk wrote:
x2suresh wrote:
arjtryarjtry wrote:
In a six-digit integer N F(k) is the value of the k-th digit. For example, F(4) is the value of the hundreds digit of N. Is N divisible by 7?

1. F(1) = F(4), F(2) = F(5), F(3) = F(6)
2. F(1) = F(2) = ... = F(6)


ABABAB
1) 100000*A +10000*B+ 1000*A +100*B + 10A +B
= 101010*A+10101* B
= (2a+b) * 10101
10101 --> is divisiable by 7

To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number

1010-2 --> 108 is divisable by 7
Suffcieint
2) A* 111111

111111 -->11111-2 =11109
11109 -->1110-18 = 1092 =
1092 --> 105 is divisable by 7

sufficient

Any other short cuts.


Am I reading this wrong? Is the number ABABAB or ABCABC? Obviously, your method would still be able to test:

ABCABC = 100000A + 10000B + 1000C + 100A + 10B + C = 100100A + 10010B + 1001C
= 1001*(100A + 10B + C)

1001/7 = 143 --> sufficient


It should be ABCABC ... Sorry I did mistake.

I modified my origianl post.

Thanks
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Re: something different..   [#permalink] 29 Aug 2008, 08:42
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In a six-digit integer N F(k) is the value of the k-th

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